Intermediate Algebra for College Students (7th Edition)

by
Blitzer, Robert F.

Published by
Pearson

ISBN 10:
0-13417-894-7

ISBN 13:
978-0-13417-894-3

Chapter 11 - Section 11.2 - Arithmetic Sequences - Exercise Set: 35

Answer

$S_{20} = 1220$

Work Step by Step

RECALL:
(1) The sum of the first $n$ terms, $s_n$, of an arithmetic sequence can be found using the formula
$S_n = \frac{n}{2}(a_1+a_n)$
where
$a_1$=first term
$a_n$ = $n^{th}$ term
(2) The $n^{th}$ term, $a_n$, of an arithmetic sequence can be found using the formula
$a_n=a_1 + d(n-1)$
where
$d$=common difference
$a_1$ = first term
To find the sum of the first 20 terms of the sequence, we need to find the value of $a_{20}$. However, the value of $a_{20}$ can only be found if we know the value of $d$.
The terms of the sequence increase by 6 so the common difference is $d=6$.
The first term of the sequence is $4$ so $a-1=4$.
Substitute these values into the formula in (2) above to obtain;
$a_n= 4+6(n-1)$
Solve for the 20th term of the sequence to obtain:
$a_{20} = 4 + 6(20-1)
\\a_{20} = 4 + 6(19)
\\a_{20} = 4 + 114
\\a_{20} = 118$
Solve for the sum of the first 20 terms using the formula in (1) above to obtain:
$S_{20} = \frac{20}{2}(4+118)
\\S_{20} = 10(122)
\\S_{20} = 1220$